Response Of 1st Order Rl And

Class Announcement Š No Class on Wed March 17 Š Help Session at 7pm, Today March 15, Bryan 305 Š Make up Class at 9am, Saturday March 20 in Bryan 305

Chapter 7. Response of 1st Order RL and RC Circuits Natural Response Response (current and voltage) of a circuit due to the initial conditions only. We will be looking at single time-constant circuits, i.e., circuits whose storage elements can be reduced to a single, equivalent storage element and whose resistors can be reduced to a single to a single, equivalent resistance.

RL Circuit – Natural Response t =0

I

R0

− vL

L

R

+ iL

The switch opens at t=0, at t=0 -, circuit is in steady state. di − i = I t = 0 At , L because dt = 0 (steady state with a DC source).

di vL = L = 0 dt iR0 = iR = 0

At t = 0, two separate circuits.

R0

I

LHS: R0 provides a path for I .

By KVL

i

I

−

+

VL

v

+

−

di di + Ri = 0 ⇒ L = − Ri ⇒ L di = − Ri dt dt dt di R = − dt i L i ( t ) dx R t = − ∫i(t0 ) x L ∫t0 dy R i (t ) ln x i (t ) = − (t − t0 ) 0 L i(t ) R =− t With t0 = 0 ⇒ ln i(0) L L

R

RHS : This is the circuit of interest.

R R − t − t i(t ) L = e ⇒ i(t ) = i(0)e L i(0) Since the current cannot change instantaneously

in an inductor, we have i(0) = i(0− ) = I . i(t ) = I ⋅ e

R − t L

−

i (t ) = I ⋅ e Let

R t L

I = I 0 . Then, i ( t ) = I ⋅ e

−

R t L

,

for t ≥ 0 .

t

τ 2τ

i(t)

.

t

5τ . .

by τ =

. 10 τ

I0

The time Unit

constant τ is given

check : τ =

Finally,

i (t ) = I ⋅ e

L v / di / dt = = sec R v/i −

t

τ

,

for t ≥ 0 .

L . R

e − t /τ . 37 . 13 . . 006 . . . . 00045

Time Constant, τ The time constant is a measure of the initial rate of change:

i (t ) = I 0 ⋅ e − t /τ 1 di = I 0 ⋅ e −t /τ ( − ) τ dt 1 di I = I 0 (− ) = − o τ τ dt t = 0 With initial value = I 0 ,

I0

i(t)

Slope (initial rate of charge) =

Linear

approximat ion to

− I 0 /τ

i ( t ) at t = 0 : −

For i = 0 , the approximat ion indicates

t=τ

t

I0

t + I0

τ t =τ.

t  −  v = i ⋅ R =  I 0e τ  ⋅ R     2t   − 2 p = i2 ⋅ R =  I0 e τ  ⋅ R     t

t

w = ∫0 p dx = ∫0 I 02 ⋅ R ⋅ e − 2 x / τ dx t

w = I 0 ⋅ R ⋅ ∫0 e − 2 x / τ dx = I 02 ⋅ R ⋅ ( 2

w = I o2 ⋅ R ⋅

τ

(

⋅ 1 − e − 2t /τ

2 L Since τ = , R

−τ )⋅e 2

−2 x t

τ 0

)

w = I o2 ⋅

(

)

L ⋅ 1 − e − 2 tR / L . 2

These relationships form the transient response of the circuit

RC Circuit – Natural Response R1

i vg

+ -

+ vR 1 -

+ vc

t =0 C

R

−

At t = 0 , the switch removes the voltage source from the capacitor. At t = 0−, circuit is in steady state. At steady state, vg is constant. Therefore,

By KVL,

dvc dv = 0. ⇒ i = c c = 0 ⇒ vR1 = 0. dt dt

vc = vg

At t=0

vg

R1

+ vg

+ -

−

Disconnected. Not doing anything.

Define a fictitious node v.

ic

v

•

+ vg

C

iR + vc −

−

ic + iR = 0

R

R C

Circuit of Interest

dv v C + =0 dt R dv v =− C dt R dv v =− dt RC dv 1 =− dt v RC dv 1 = − ∫ v RC ∫ dt v ( t ) dy 1 t = − ∫v (0) y RC ∫o dx

t RC v(t ) t ln =− v (o ) RC v (t )

ln y v ( o ) = −

t

− v(t ) = e RC v (o )

t≥0

v(0 − ) = v(0) = v g =v I v(t ) = vI e −t / RC Set τ = RC. v(t ) = vI e −t / τ

t≥0

Step Response of First Order RL and RC Circuits

A step response is the response of a circuit (voltage/current) to the sudden application of a constant voltage or current source. Consider:

vs

t=0

R +i v

+ −

At t = 0 − ,

L

−

i (0 − ) = I 0

- initial inductor current

Rs

I0

For t > 0: R Vs

+

+

i

−

KVL

v −

− v s + iR + v = 0 − v s + iR + L

di =0 dt

di = vs dt di R v + i= s dt L L

iR + L

This is a standard form ODE.

L

i (t )

v di R + i= s L dt L v di R = s − i dt L L di R v = ( s − i) dt L R R v di = ( s − i ) dt L R di R = dt v L ( s − i) R dy R t i (t ) = ∫i ( 0 ) v ∫0 L s − y R i (0) = I 0

t

vs R − ln( − y ) = x R L 0 I0 − ln(

dx

v vs R − i ) + ln( s − I 0 ) = t L R R

vs −i R R =− t ln vs L − I0 R vs R −i − t L R =e vs − I0 R −R t vs vs L − i = ( − I0 ) ⋅ e R R

R

− t vs vs L + (I0 − ) ⋅ e i (t ) = R R v (t ) = v s − i (t ) R − t  vs vs L  v (t ) = v s − R  + ( I 0 − ) ⋅ e  R R  R − t vs L v (t ) = ( − I 0 ) R ⋅ e R R

Other Approaches R Vs

v(t )

+

+ −

i

L

v −

Vs R

R i2

i2 = By KCL : v (t ) − v s + iL = 0 R

vs R

di1 =0 dt v di (i1 − s ) R + L 1 = 0 R dt di1 =0 i1 R − vs + L dt (i1 − i2 ) R + L

i1

+ L −

Example Rs

R Vs

+ -

+

i v −

t =0 C

R2

+ -

Solve for v(t) and i(t). First note that t = 0− , v(0− ) = V0

R2 . R2 + Rs

V0

For t ≥ 0 : R Vs

+ −

v − Vs + ic = 0 R v − Vs dv +C =0 R dt dv Vs − v = dt RC V −v dv = s dt RC v ( t ) dy 1 t ∫v(0) Vs − y = RC ∫0 dx

C

+ v

v(0) = V0

−

v (t )

− ln(Vs − y ) v (0) =

R2 ∆ VI R2 + Rs

t RC

ln(Vs − v(t )) − ln(Vs − VI ) = − ln

t RC

Vs − v(t ) − t = Vs − VI RC

Vs − v(t ) = e −t / RC Vs − VI Vs − v(t ) = (Vs − VI )e −t / RC − v(t ) = −Vs + (Vs − VI )e −t / RC v(t ) = Vs + (VI − Vs )e −t / RC

General Form t 1 dx = ∫x(t0 ) x f − x ∫t0 τ dt x (t )

dx x + =K dt τ at t → ∞

xf dx → 0, = K, xf = τ ⋅ K τ dt

x dx =K− τ dt dx 1 = ( Kτ − x) dt τ 1 dx = ( Kτ − x) dt

τ

1 dx = ( x f − x) dt

τ

1 dx = dt x f −x τ

− ln( x f − x)

x (t ) x ( t0 )

=

t

τ

t

t0

− ln( x f − x(t )) + ln( x f − x(t0 )) = ln

x f − x(t ) x f − x(t0 )

x f − x(t ) x f − x(t0 )

=

t − t0

− (t − t0 )

τ

= e − ( t −t 0 ) / τ

[

]

x f − x(t ) = x f − x(t0 ) ⋅ e −(t −t0 ) /τ

[

]

− x(t ) = − x f + x f − x(t0 ) ⋅ e −(t −t0 ) /τ

[

]

x(t ) = x f + x(t0 ) − x f ⋅ e −(t −t0 ) /τ

τ

General Solutions

The unknown variable as = a function of t

⇒

The final value of + the variable

The initial The final value of - value of the variable the variable

x ( t ) = x final + [x initial − x final ]⋅ e − t / τ

× e −t /(time cons tan t )

Return to Previous Example R Vs

i

Solution Method 1 V v − s + +i = 0 R R dv i =C dt V v dv Vs − s + +C = 0 dt R R R V dv v + = s dt RC RC Solve this ODE.

+ v

i

+ −

C

−

vinitial = v(0) = VI V −V iinitial = i(0) = s I R

Source Transform

Solution Method 2

R

i

+

C

v −

ifinal = C

dv =0 dt t =∞

vfinal = Vs V −V i(t ) = s I e −t / RC R

v(t ) = Vs + (VI − Vs )e −t / RC

Sequential Switching Circuits may have more than one switch, opening and closing at different times. Such a circuit is analyzed in stages for different time periods. It is important to realize that the initial conditions for one time period are derived from the previous time period. Note: iL and vc cannot change instantaneously.

4Ω

+ -

60V

t = 35ms

t =0 3Ω

12Ω

6Ω

For t ≥ 0, find iL(t).

iL

+ vL −

150×10−3 H

18Ω

At t = 0 − , v L = L

diL = 0 ⇒ The inductor is like a short circuit. dt 3Ω

4Ω

+ -

6Ω

12Ω

60V

iL

3Ω

60 A 4

4Ω

12Ω

1 1 1 1 3 +1+ 2 1 = + + = = 12 2 Req 4 12 6

6Ω

⇒ Req = 2Ω

3Ω

15A

iL

Current Divider iL = 15

2Ω

iL

2 2 = 15 = 6 A 2+3 5

iL (0) = 6 A

Consider 0 ≤ t ≤ 35×10−3 (t0 = 0). 3Ω

iL

6Ω

9Ω

150 ×10−3 H

18Ω

18Ω

iL

iLf = 0

1 1 3 1 + = = ⇒ Req = 6Ω 9 18 18 6

6Ω

150 × 10 −3 H

τ=

L 0.15 1 = = R 6 40

iL (t ) = iLf + (iL (0) − iLf

iL

−(t −0) )⋅e L/ R −3

−3

150 × 10 H

= 6 ⋅ e−40t

iL (t = 35×10−3 ) = 6 ⋅ e−40(35×10 ) = 6 ⋅ e−1.4 = 1.48A

Consider t ≥ 35×10−3

(t0 = 0.035). i LI = 1 .48 A

3Ω

iL f = 0 A 6Ω

iL

0.15H

L 0.15 1 = = R 9 60 i L (t ) = 0 + (1.48 − 0) ⋅ e − ( t − 0.035 )⋅60 = 1 .48 ⋅ e − ( t − 0.035 )⋅60

τ=

iL (t )

τ=

6

1 40

τ=

1 60

1 . 48

0

35 x 10 −3

t

Unbounded Response Consider the following circuit. The initial voltage across the capacitor v(0-) = 10V.

+ 10V −

+ 5µF

v

t =0 10 kΩ

4i∆

i∆

20 kΩ

−

At t=0, the switch closes, allowing the capacitor to begin discharging through the resistors. This produces an i∆ . The i∆ controls the current in the dependent current source.

Thevenize

Need

V t (i.e., open circuit voltage) and Rt looking through the two terminals.

Without any external source, i∆ = 0. ⇒ Voc = Vt = 0.

Rt is best determined by a test involving an application of a certain source. Apply a test voltage vT, and measure the test current iT. iT

vT

+ _

10 kΩ

4i∆

i∆

20 kΩ

vT v v − 4i∆ + T = 0 and i∆ = T . 10 k 20 k 20 k v v 4v − iT + T − T + T = 0 10 k 20 k 20 k 1 v vT ( − ) = iT ⇒ T = −20 kΩ = Rt 20 k iT − iT +

We now have the following equivalent Thevenin circuit. t =0 + 10V −

− 20kΩ

5µF

Note: Negative resistance. What does negative resistance mean? Ohms Law : v = iR If

R

is negative, v = i(−R) = −iR < 0.

⇒ The resistor behaves like a source.

Solves for v(t), the voltage across the capacitor:

dv v + =0 dt R dv v =− dt RC dv 1 =− dt v RC .... c

v(t ) = VI e

−

1 t RC

v (t ) = 10 ⋅ e − t /( −20 , 000 ⋅5⋅10 v (t ) = 10 ⋅ e +10 t , The final value is unbound.

−6

)

t≥0

v

t

Example Initial capacitor voltages

t = 0

− 5V +

6µF i +

−

1µ F

30 V

2µF

250 k

+ v0

−

2µF

1 µ F = 3 µ F with

Then the The

resulting

3µF

capacitor overall

initial is capacitor

voltage in

of series

is

30 V . with

3⋅6 = 2µF . 3 + 6

6µF .

t=0

+ 25 V −

+ vo

i

2 µF

v oI = 25V

250 k Ω

−

and

V of = 0V .

v ( t ) = v oI e − t / RC = 25 e − t / 2 ⋅10

−6

⋅ 250 ⋅10 3

v ( t ) = 25 e − 2 t To find the voltage across each capacitor:

v(t ) 25 ⋅ e −2t i (t ) = = = 10 − 4 ⋅ e −2t 250,000 R 1t v = ∫ i ( x) dx + initial voltage of the capacitor C0

− 10 − 4 ⋅ e − 2 t

+

6µF

v1

+ v

−

3µF

+ v2

250 k Ω

−

−

v1 (0) = −5V and v2 (0) = 30V . 1 t −4 − 2t v1 (t ) = − 10 ⋅ e dx + (−5) ∫ −6 0 6(10 )

−1 t −4 − 2t v 2 (t ) = 10 ⋅ e + 30 − 6 ∫0 3(10 )

v1 (t ) = −16.67 ∫0 e − 2 x dx − 5

v 2 ( t ) = − 16 . 67 ⋅ (1 − e − 2 t ) + 30

v1 (t ) = −8.33 ⋅ (1 − e − 2t ) − 5

v 2 ( t ) = 13 . 33 + 16 . 67 ⋅ e − 2 t

v1 (t ) = −13.33 + 8.33 ⋅ e −2t

lim v ( t ) = 13 .33V

t

lim v (t ) = −13.33V t →∞

1

t→∞

2

p = i2R

(

−4

⋅e

(

−4

p = 10 ∞ 0

w = ∫ 10 Initial

)

−2t 2

⋅e

dw 250 , 000 = dt

) 250 ,000 dt = 625 µ J

−2t 2

Energy

1 1 1 CV 2 = ⋅ 6 ⋅ 10 − 6 ( 25 ) + ( 2 + 1)(10 − 6 )( 30 2 ) = 1, 425 µ J 2 2 2 Final Energy 1 1 = 6 ⋅ 10 − 6 (13 . 33 ) 2 + ( 3 )(10 − 6 )(13 . 33 2 ) ≅ 800 µ J 2 2

=